Quadratic Equations

A quadratic equation is any equation where the highest power of the variable is 2. The standard form is ax² + bx + c = 0 where a ≠ 0. Think of it like this — linear equations give us straight lines, quadratics give us parabolas (U-shaped curves). And where that parabola crosses the x-axis? Those are our roots (solutions).

Methods to Solve

We have three weapons:

1. Factorization — fastest when it works
2. Quadratic formula — works every single time
3. Completing the square — rarely needed in exams, but good to know

Method 1: Factorization

The idea is to split the middle term bx into two parts that let us factor nicely.

Steps:

1. Multiply a × c
2. Find two numbers that multiply to give a×c and add up to b
3. Split the middle term and factor by grouping

Example 1: Simple factorization

Solve: x² - 7x + 12 = 0

We need two numbers that multiply to 12 and add to -7. That’s -3 and -4.

• x² - 3x - 4x + 12 = 0
• x(x - 3) - 4(x - 3) = 0
• (x - 3)(x - 4) = 0
• x = 3 or x = 4

Shortcut: For x² + bx + c = 0 (when a = 1), the roots are just the two numbers that multiply to c and add to b. We can skip the grouping entirely.

Example 2: When a ≠ 1

Solve: 2x² + 7x + 3 = 0

a × c = 2 × 3 = 6. We need two numbers multiplying to 6 and adding to 7. That’s 6 and 1.

• 2x² + 6x + x + 3 = 0
• 2x(x + 3) + 1(x + 3) = 0
• (2x + 1)(x + 3) = 0
• x = -1/2 or x = -3

Method 2: Quadratic Formula

When factorization isn’t clean (ugly numbers, irrational roots), this is our go-to.

Example 3: Using the formula

Solve: 3x² - 5x + 1 = 0

Here a = 3, b = -5, c = 1.

D = b² - 4ac = 25 - 12 = 13

x = (5 ± √13) / 6

x = (5 + √13)/6 or x = (5 - √13)/6

The Discriminant — Nature of Roots

The discriminant D = b² - 4ac tells us everything about the roots without actually solving.

Example 4: Finding nature of roots

For what values of k does 2x² + kx + 8 = 0 have equal roots?

For equal roots, D = 0:

• k² - 4(2)(8) = 0
• k² = 64
• k = ±8

Sum and Product of Roots

This is a massive shortcut. We can find relationships between roots without actually solving the equation.

If α and β are roots of ax² + bx + c = 0:

• Sum: α + β = -b/a
• Product: αβ = c/a

Example 5: Using sum and product

If the roots of x² - 5x + 6 = 0 are α and β, find α² + β².

Sum = α + β = 5, Product = αβ = 6.

Now the trick: α² + β² = (α + β)² - 2αβ = 25 - 12 = 13

This identity (α² + β²) = (α + β)² - 2αβ comes up ALL the time. Memorize it.

Other useful identities:

• α² + β² = (α + β)² - 2αβ
• (α - β)² = (α + β)² - 4αβ
• α³ + β³ = (α + β)³ - 3αβ(α + β)

Forming Equations from Roots

If we know the roots, we can build the equation: x² - (sum of roots)x + (product of roots) = 0

Example 6: Building an equation

Form a quadratic equation whose roots are 2 and -5.

Sum = 2 + (-5) = -3. Product = 2 × (-5) = -10.

Equation: x² - (-3)x + (-10) = 0 → x² + 3x - 10 = 0

The Sign-Based Comparison Trick

This shows up in banking exams constantly. We get two equations and need to compare x and y.

The method:

1. Solve both equations to find roots of x and roots of y.
2. Compare the roots directly.

In simple language, we just need to figure out which variable has larger values. If all roots of x are greater than all roots of y, then x > y. If there’s overlap, the relationship can’t be determined.

Example 7: Comparison problem

I: x² - 7x + 12 = 0 II: y² - 9y + 20 = 0

Find the relationship between x and y.

Equation I: (x-3)(x-4) = 0 → x = 3 or 4 Equation II: (y-4)(y-5) = 0 → y = 4 or 5

Comparing: x can be 3 or 4, y can be 4 or 5. Since x values (3, 4) are ≤ y values (4, 5):

x ≤ y

Common Exam Variations

• Find the value of k such that the equation has equal roots / no real roots / one root is twice the other.
• Comparison of roots from two different equations (banking exam favorite).
• Sum/product of roots without solving — “find α² + β²” or “find 1/α + 1/β”.
• If one root is known, find the other root and the unknown coefficient.
• Common root between two quadratic equations.

Shortcut: Finding 1/α + 1/β

Instead of solving for individual roots:

1/α + 1/β = (α + β)/(αβ) = (-b/a)/(c/a) = -b/c

Super fast for exam questions!

Practice Problems

Problem 1: Solve 6x² - x - 2 = 0 by factorization.

Problem 2: If the roots of 2x² - 8x + k = 0 are equal, find k.

Problem 3: The roots of x² - px + 12 = 0 are in the ratio 1:3. Find p.

Answers

Problem 1: a×c = -12. Numbers: -4 and 3 (multiply to -12, add to -1). 6x² - 4x + 3x - 2 = 0 → 2x(3x - 2) + 1(3x - 2) = 0 → (2x + 1)(3x - 2) = 0. x = -1/2 or x = 2/3.

Problem 2: For equal roots, D = 0. b² - 4ac = 64 - 8k = 0 → k = 8.

Problem 3: Let roots be α and 3α. Sum: α + 3α = 4α = p → α = p/4. Product: α × 3α = 3α² = 12 → α² = 4 → α = 2. So p = 4α = 4(2) = 8. (We take α = 2 since ratio is 1:3, both positive.)